If the angle between the asymptotes of hyperbola a 2 x 2 − b 2 y 2 = 1 is 3 π . Then the eccentricity of conjugate hyperbola is

2 tan − 1 ( a b ) = 3 π a b = 3 1 e 2 = 1 + 3 1 = 3 4 e ′ 2 1 + e 2 1 = 1 ⇒ e ′ 2 1 + 4 3 = 1 ⇒ e ′ 2 1 = 4 1 ⇒ e ′ = 2

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MathematicsParabolaMiscellaneous/MixedMedium2 minQB

MathematicsMediuminteger

If the angle between the asymptotes of hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is $\frac{π}{3}$ . Then the eccentricity of conjugate hyperbola is

Answer:

Solution:

$2 tan^{- 1} (\frac{b}{a}) = \frac{π}{3}$ $\frac{b}{a} = \frac{1}{3}$ $e^{2} = 1 + \frac{1}{3} = \frac{4}{3}$ $\frac{1}{e ^{'^{2}}} + \frac{1}{e ^{2}} = 1$ $\Rightarrow \frac{1}{e ^{'^{2}}} + \frac{3}{4} = 1$ $\Rightarrow \frac{1}{e ^{'^{2}}} = \frac{1}{4} \Rightarrow e^{'} = 2$

Stream:JEESubject:MathematicsTopic:ParabolaSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: QB

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